79. Word Search #

题目 #

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

题目大意 #

给定一个二维网格和一个单词，找出该单词是否存在于网格中。单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

解题思路 #

在地图上的任意一个起点开始，向 4 个方向分别 DFS 搜索，直到所有的单词字母都找到了就输出 true，否则输出 false。

代码 #


package leetcode

var dir = [][]int{
	[]int{-1, 0},
	[]int{0, 1},
	[]int{1, 0},
	[]int{0, -1},
}

func exist(board [][]byte, word string) bool {
	visited := make([][]bool, len(board))
	for i := 0; i < len(visited); i++ {
		visited[i] = make([]bool, len(board[0]))
	}
	for i, v := range board {
		for j := range v {
			if searchWord(board, visited, word, 0, i, j) {
				return true
			}
		}
	}
	return false
}

func isInBoard(board [][]byte, x, y int) bool {
	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
}

func searchWord(board [][]byte, visited [][]bool, word string, index, x, y int) bool {
	if index == len(word)-1 {
		return board[x][y] == word[index]
	}
	if board[x][y] == word[index] {
		visited[x][y] = true
		for i := 0; i < 4; i++ {
			nx := x + dir[i][0]
			ny := y + dir[i][1]
			if isInBoard(board, nx, ny) && !visited[nx][ny] && searchWord(board, visited, word, index+1, nx, ny) {
				return true
			}
		}
		visited[x][y] = false
	}
	return false
}

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